LeetCode 3666 - Minimum Operations to Equalize Binary String

Difficulty: 🔴 Hard
Topics: Math, String, Breadth-First Search, Union-Find, Ordered Set

Solution

Problem Understanding

The problem asks us to transform a binary string s consisting of characters '0' and '1' into a string where all characters are '1'. The allowed operation is flipping exactly k different indices in one move, turning '0' to '1' and '1' to '0'. The goal is to find the minimum number of such operations needed to achieve the all-ones string, or return -1 if it is impossible.

The input consists of a string s with length up to 10^5 and an integer k between 1 and the length of s. This large size hints that any brute-force approach that attempts to simulate all possible combinations of flips is infeasible. Important edge cases include situations where the number of '0's is less than k, or where k is exactly 1, which simplifies the problem considerably. Another tricky scenario is when k is even, as flipping an even number of bits can create situations where some '0's cannot be isolated without affecting previously corrected positions.

The problem guarantees that s only contains '0' and '1' characters, and that k is always valid with respect to the string length.

Approaches

Brute Force

A brute-force approach would attempt to generate all possible combinations of k indices to flip in each operation. For each possible combination, it would flip the selected bits and recursively try further flips until the string becomes all ones. While this guarantees correctness, it is extremely inefficient because the number of combinations grows combinatorially with the string length and k. This approach is completely impractical for strings of length 10^5.

Optimal Approach

The key observation is that the operation can be seen as a sliding "window" of influence. If we process the string from left to right, we can track how many flips affect the current bit using a difference array or prefix sum approach. Each time we encounter a '0' that has been flipped an even number of times (effectively still '0'), we need to flip the next k bits starting from that position. Using this approach, we can compute the minimum number of flips in linear time, provided it is possible. If at any point we cannot flip k consecutive bits because it would go beyond the string boundary, we return -1.

This transforms the problem into a linear pass with a queue or difference array to efficiently track the net flips affecting each bit.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(2^n)	O(n)	Try all combinations of `k` indices to flip; infeasible for large `n`.
Optimal	O(n)	O(n)	Linear scan using difference array or prefix sum to track flips; efficiently computes minimal operations.

Algorithm Walkthrough

Initialize a variable flip_count to track the number of operations performed.
Use an auxiliary array diff of size n+1 to track how flips affect subsequent indices. The value at diff[i] represents the net effect of flips starting at index i.
Initialize current_flips to 0, representing the number of flips affecting the current index.
Iterate over the string s from left to right.
At each index i, update current_flips by adding diff[i] to account for flips ending at this position.
Check if the current bit, after accounting for flips (current_flips % 2), is '0'. If it is:
If there are fewer than k bits remaining starting from index i, return -1.
Otherwise, increment flip_count by 1.
Increment current_flips by 1 to indicate the new flip starting at index i.
Decrement diff[i+k] by 1 to indicate the flip effect ends after k positions.
After finishing the iteration, return flip_count.

Why it works: By tracking the net flips affecting each position, we ensure that every '0' is flipped at the earliest possible index using exactly k flips. The difference array ensures that previous flips' influence is efficiently propagated, guaranteeing a minimal number of operations.

Python Solution

class Solution:
    def minOperations(self, s: str, k: int) -> int:
        n = len(s)
        diff = [0] * (n + 1)
        flip_count = 0
        current_flips = 0

        for i in range(n):
            current_flips += diff[i]
            bit = int(s[i]) ^ (current_flips % 2)
            if bit == 0:
                if i + k > n:
                    return -1
                flip_count += 1
                current_flips += 1
                diff[i + k] -= 1

        return flip_count

This implementation uses a linear scan with a difference array to track flips efficiently. current_flips % 2 determines the effective value of the current bit after all prior flips. Whenever a '0' is encountered, we perform a flip operation over the next k bits, increment flip_count, and schedule the end of the flip effect in diff.

Go Solution

func minOperations(s string, k int) int {
    n := len(s)
    diff := make([]int, n+1)
    flipCount := 0
    currentFlips := 0

    for i := 0; i < n; i++ {
        currentFlips += diff[i]
        bit := int(s[i]-'0') ^ (currentFlips % 2)
        if bit == 0 {
            if i+k > n {
                return -1
            }
            flipCount++
            currentFlips++
            diff[i+k]--
        }
    }

    return flipCount
}

The Go implementation mirrors the Python solution. Differences include explicit conversion from byte to int when computing the bit value, and slice initialization using make. Otherwise, the logic and efficiency are identical.

Worked Examples

Example 1: s = "110", k = 1

i	s[i]	effective bit	Action	flip_count	diff
0	1	1	-	0	[0,0,0,0]
1	1	1	-	0	[0,0,0,0]
2	0	0	flip	1	[0,0,1,-1]

Output: 1

Example 2: s = "0101", k = 3

i	s[i]	current_flips	effective bit	Action	flip_count	diff
0	0	0	0	flip	1	[0,0,0,-1,0]
1	1	1	0	flip	2	[0,1,0,-1,-1]
2	0	1	1	-	2	[0,1,0,-1,-1]
3	1	0	1	-	2	[0,1,0,-1,-1]

Output: 2

Example 3: s = "101", k = 2

i	s[i]	current_flips	effective bit	Action	flip_count	diff
0	1	0	1	-	0	[0,0,0,0]
1	0	0	0	flip	1	[0,0,1,-1]
2	1	1	0	cannot flip beyond n	-	-

Output: -1

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Single linear pass over the string, updating `current_flips` and `diff`.
Space	O(n)	Difference array of size `n+1` to track flips efficiently.

The complexity is linear because each index is visited exactly once, and the flip effect propagation is handled in constant time using the difference array.

Test Cases

# Provided examples
assert Solution().minOperations("110", 1) == 1  # single zero flip
assert Solution().minOperations("0101", 3) == 2  # multiple flips needed
assert Solution().minOperations("101", 2) == -1  # impossible

# Edge cases
assert Solution().minOperations("1", 1) == 0  # already all ones
assert Solution().minOperations("0", 1) ==